Q: A committee of 4 is to be formed from 7 men and 5 women. How many ways can the committee be formed if it must contain exactly 2 women?

$$\displaystyle 5C2 \times 7C2$$

Q: In a video game, Ben has 15 archers, 10 swordsmen, and 5 knights. He wants to send 5 archers, 3 swordsmen, and 2 knights to the arena. How many ways can he form his team?

$$15C5 \times 10C3 \times 5C2$$

Q: How many ways can you select a committee of 5 people from 5 graduate students and 7 undergraduates, if at least 3 must be undergraduates?

$$7C3 \times 5C2 + 7C4 \times 5C1 + 7C5$$

Q: How many ways can you arrange the letters in the word 'PERMUTATIONS' so that S is always to the right of P?

$$\displaystyle \frac{13!}{2!} \times \frac{1}{2}$$

Question 1

Suppose you want to create a 6-digit password using the digits 0-9, where digits can be repeated. What is the total number of possible passwords?

Accepted Answer

$$10^6$$

Question 2

A committee of 4 is to be formed from 7 men and 5 women. How many ways can the committee be formed if it must contain exactly 2 women?

Accepted Answer

$$\displaystyle 5C2 	imes 7C2$$

Question 3

How many different ways can the letters in the word 'TENNESSEE' be arranged?

Accepted Answer

$$\displaystyle \frac{9!}{2!4!2!1!}$$

Question 4

A round table has 8 seats. In how many ways can 8 people be seated if rotations are considered the same arrangement?

Accepted Answer

$$7!$$

Question 5

You have 4 different coats, 3 different jeans, and 2 different pairs of shoes. If you must wear one of each, how many different outfits can you make?

Accepted Answer

$$4 	imes 3 	imes 2 = 24$$

Question 6

How many ways can 5 officers be assigned to patrol, 2 to paperwork, and 3 to the donut shop from a group of 10 officers?

Accepted Answer

$$\displaystyle \frac{10!}{5!2!3!}$$

Question 7

A password consists of 3 letters followed by 2 digits. Letters and digits can be repeated. How many different passwords are possible?

Accepted Answer

$$26^3$$ 	imes $$10^2$$

Question 8

How many ways can 6 boys and 5 girls be seated in a row so that no two girls are together?

Accepted Answer

$$6! 	imes 7P5$$

Question 9

In a video game, Ben has 15 archers, 10 swordsmen, and 5 knights. He wants to send 5 archers, 3 swordsmen, and 2 knights to the arena. How many ways can he form his team?

Accepted Answer

$$15C5 	imes 10C3 	imes 5C2$$

Question 10

How many different 4-letter words can be formed using 3 A's, 5 B's, and 7 C's?

Accepted Answer

$$\displaystyle \frac{4!}{a!b!c!}$$,\ a+b+c=4$

Question 11

A deck of 26 cards is to be split evenly between two players. How many ways can this be done?

Accepted Answer

$$\displaystyle \frac{26!}{13!13!}$$

Question 12

How many ways can the letters in the word 'EDUCATION' be arranged if the letters A and T must always be together?

Accepted Answer

$$8! 	imes 2!$$

Question 13

Suppose you have 12 students and 12 different versions of an exam. In how many ways can you distribute the exams so that each student gets a different version?

Accepted Answer

$$12!$$

Question 14

How many ways can you arrange the books if you have 2 physics, 4 chemistry, 2 botany, and 3 zoology books, and books of the same subject must stay together?

Accepted Answer

Question 15

How many ways can you select a committee of 5 people from 5 graduate students and 7 undergraduates, if at least 3 must be undergraduates?

Accepted Answer

$$7C3 	imes 5C2 + 7C4 	imes 5C1 + 7C5$$

Question 16

How many ways can you arrange the letters in the word 'PERMUTATIONS' so that S is always to the right of P?

Accepted Answer

$$\displaystyle \frac{13!}{2!} 	imes \frac{1}{2}$$

Question 17

Suppose you have a password of length 10, with 2 '1's, 3 '2's, 3 'A's, and 2 '!'. How many different passwords can you form?

Accepted Answer

$$\displaystyle \frac{10!}{2!3!3!2!}$$

Question 18

How many ways can 10 students sign up for 20 sports, if each student signs up for one sport and at least one pair of students must sign up for the same sport?

Accepted Answer

Question 19

How many ways can you assign 6 workers to 3 jobs (A, B, C) if A needs 2 workers, B needs 3, and C needs 1?

Accepted Answer

$$\displaystyle \frac{6!}{2!3!1!}$$

Combinatorial Probability and Counting Techniques in Statistics

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